Question

Got a Calculus question regarding antiderivatives, screenshot in the comments! <3 <3 <3

Answer 1

\[\text{ say we have } h(x)=\int\limits_{a(x)}^{b(x)}f(t) dt\]
And say you knew F'=f then evaluate the integral I gave you

Answer 2

\[h(x)=\int\limits_{a(x)}^{b(x)}f(t) dt=F(t)|_{t=a(x)}^{t=b(x)}=F(b(x))-F(a(x)) \\ \text{ so you have } h(x)=F(b(x))-F(a(x)) \\ \text{ you can use chain rule to find h' }\]

Answer 3

What would be h(x) in this case? Kinda confused about what to do with this information.

Answer 4

Do you know the chain rule?

Answer 5

\[h'(x)=b'(x) \cdot f(b(x))-a'(x) \cdot f(a(x))\]
by chain rule

Answer 6

you know what f(t) is here

Answer 7

so what is f(b(x))
your b(x) is sin(x)

Answer 8

and your a(x) is -4

Answer 9

I basically gave you a formula to use

Answer 10

I know you gave the formula, thank you, was just kinda confused on what to do with it.

Answer 11

you just need to identify what f(t) is
a(x) is
b(x) is
and find a'(x),b'(x), f(b(x)), and f(a(x))

Answer 12

Trying to read over your info now again.

Answer 13

let me know if you don't understand something

Answer 14

so f(t) is cos(t^5)+t, correct? So, I find the antiderivative of that and plug in b and a in to the antiderivative and then subtract b(x) and f(b(x) by a(x)a(b(x))?

Answer 15

you don't need to find the antiderivative of f

Answer 16

What is the integrand there for then? Maybe that's why I'm confused.

Answer 17

do you think happens when we differentiate an integral?
like for example
\[\frac{d}{dx}\int\limits_{}^{}f(x) dx =? \\ \]

Answer 18

ohhhh it stays as f(x)

Answer 19

\[\frac{d}{dx} \int\limits (2x+5) dx=\frac{d}{dx} (\frac{2x^2}{2}+5x+C)=2x+5\]
yep

Answer 20

so what I'm saying is above we don't need to actually integrate the integrand

Answer 21

but we will definitely use the integrand to answer the question

Answer 22

ahhh ok. so just plug in sinx for t and -4 for t?

Answer 23

yes plug those into the integrand
but you will also need find the derivatives of sin(x) and -4

Answer 24

\[h'(x)=b'(x)f(b(x))-a'(x)f(a(x)) \\ =(\sin(x))'[\cos(\sin(x^5))+\sin(x)]-(-4)'[\cos((-4)^5)-4]\]

Answer 25

0

Answer 26

so cos(x)[cos(sin(x^5))+sinx?]

Answer 27

looks good

Answer 28

oops one sec

Answer 29

lol I think I put that 5 in the wrong spot :p

Answer 30

\[f(t)=\cos(t^5)+t \\ f(\sin(x))=\cos(\sin^5(x))+\sin(x)\]
yea I put that 5 in the wrong spot

Answer 31

\[\cos(x)(\cos(\sin^5(x))+\sin(x))\]

Answer 32

so anyways do you kinda understand it more?

Answer 33

ya, but my homework thing said that wasn't the right answer.

Answer 34

\[\frac{d}{dx} \int\limits_{-4}^{\sin(x)}(\cos(t^5)+t) dt \]
say we know the antiderivative of cos(t^5)+t is F(t) then
\[\frac{d}{dx}(F(t)|_{t=-4}^{t=\sin(x)}) \\ \frac{d}{dx}(F(\sin(x))-F(-4)) \\ \cos(x)f(\sin(x))-0 \\ \cos(x) \cdot (\cos(\sin^5(x))+\sin(x))\]
it definitely is the answer

Answer 35

like how are you putting it in

Answer 36

you put all the ( ) needed

Answer 37

in you put the 5 in the right spot?

Answer 38

cos(x)(cos(sin5(x))+sin(x))

Answer 39

oh wiat, forgot the exponent sign

Answer 40

cos(x)(cos(sin^5(x))+sin(x))

Answer 41

ok that did it. Thanks so much! Sorry I'm such a bonehead =P

Answer 42

lol its fine
math is so sensitive

Answer 43

you hurt its feelings real easy

Answer 44

hahaha, well thanks again!

Answer 45

np